Page 71 - eEISBN FINALIZE
P. 71
Solution :
velocity-time graph
v ms -1
A B
2.8
C
0 4 17 23 t s
i.Acceleration, a = gradient of 0A
2.8 − 0
=
4 − 0
= . −2
ii. Decelaration, a = gradient of BC
0 − 2.8
=
23 − 17
= − . −2
iii. Distance during acceleration, S1= the area under the graph. (area of triangle 0-A-
4).
1
= (4)(2.8)
1
2
= .
1
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